Problem: What is the smallest positive integer that leaves a remainder of 4 when divided by 5 and a remainder of 6 when divided by 7?
Solution: Let the desired number be $a$. Then \begin{align*}
a\equiv 4\pmod 5,\\
a\equiv 6\pmod 7.
\end{align*} The first congruence means that there exists a non-negative integer $n$ such that $a=4+5n$. Substituting this into the second congruence yields \[4+5n\equiv 6\pmod 7\implies n\equiv 6\pmod 7\] So $n$ has a lower bound of $6$. Then $n\ge 6\implies a=4+5n\ge 34$. $\boxed{34}$ is the smallest solution since it is a lower bound of $a$ and satisfies both original congruences.